1. **State the problem:** Given parametric equations $x = k(t - \sin t)$ and $y = k(1 - \cos t)$ with $k \neq 0$, find the second derivative $\frac{d^2 y}{dx^2}$ at $t = \frac{\pi}{2}$. 2. **Find the first derivatives:** $$\frac{dx}{dt} = k(1 - \cos t)$$ $$\frac{dy}{dt} = k \sin t$$ 3. **Find the first derivative $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{k \sin t}{k(1 - \cos t)} = \frac{\sin t}{1 - \cos t}$$ 4. **Simplify $\frac{dy}{dx}$ using half-angle identities:** Recall that $1 - \cos t = 2 \sin^2 \frac{t}{2}$ and $\sin t = 2 \sin \frac{t}{2} \cos \frac{t}{2}$, so $$\frac{dy}{dx} = \frac{2 \sin \frac{t}{2} \cos \frac{t}{2}}{2 \sin^2 \frac{t}{2}} = \frac{\cos \frac{t}{2}}{\sin \frac{t}{2}} = \cot \frac{t}{2}$$ 5. **Find the second derivative $\frac{d^2 y}{dx^2}$:** By chain rule, $$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \cot \frac{t}{2} \right) \cdot \frac{dt}{dx}$$ 6. **Calculate $\frac{d}{dt} \left( \cot \frac{t}{2} \right)$:** $$\frac{d}{dt} \cot \frac{t}{2} = -\csc^2 \frac{t}{2} \cdot \frac{1}{2} = -\frac{1}{2} \csc^2 \frac{t}{2}$$ 7. **Calculate $\frac{dt}{dx}$:** Since $\frac{dx}{dt} = k(1 - \cos t) = 2k \sin^2 \frac{t}{2}$, $$\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{2k \sin^2 \frac{t}{2}}$$ 8. **Combine to find $\frac{d^2 y}{dx^2}$:** $$\frac{d^2 y}{dx^2} = -\frac{1}{2} \csc^2 \frac{t}{2} \cdot \frac{1}{2k \sin^2 \frac{t}{2}} = -\frac{1}{4k} \csc^2 \frac{t}{2} \cdot \csc^2 \frac{t}{2} = -\frac{1}{4k} \csc^4 \frac{t}{2}$$ 9. **Evaluate at $t = \frac{\pi}{2}$:** $$\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$$ $$\csc \frac{\pi}{4} = \frac{1}{\sin \frac{\pi}{4}} = \sqrt{2}$$ Therefore, $$\csc^4 \frac{\pi}{4} = (\sqrt{2})^4 = 4$$ 10. **Final value:** $$\frac{d^2 y}{dx^2} \bigg|_{t=\frac{\pi}{2}} = -\frac{1}{4k} \times 4 = -\frac{1}{k}$$ **Answer:** $\boxed{-\frac{1}{k}}$ which corresponds to option (a).